(07/20/2010, 08:31 PM)tommy1729 Wrote: how do you know the superfunction of 2sinh is periodic in the imaginary direction ?

that is intresting.

It is interesting. The period is based on the limiting behavior of 2sinh at its fixed point of zero, where the slope=2. If -real(z) is large enough, then 2^z becomes an excellent approximation of the SuperFunction of 2sinh. The limit equation for the superfunction of 2sinh in the complex plane is:
Since 2^z is periodic in the imaginary, with period = i*2pi/ln(2), then the SuperFunction of 2sinh is also periodic with that period. This leads to some interesting behavior (pointed out in earlier posts). The SuperFunction grows superexponentially negative at img(z)=i*pi/ln(2), with img(f)=0. At img(z)=0.5i*pi/ln(2), the SuperFunction has real(f)=0, and converges to an imaginary fixed point.
- Sheldon

since a^z is periodic in direction Q , with period i*2pi/ln(a) , then the superfunction of af is also periodic with that period.

Mostly correct, with period i*2pi/ln(a). But not true for every superfunction. Knesser's solution for sexp_e is not periodic, but is pseudo periodic. It starts with the superfunction, which is complex valued at the real axis, and then does a conformal mapping, to put real values back on the real axis, along with a Schwarz reflection. For i(z)>=1i, sexp_e is fairly close to the (linearly shifted) complex valued superfunction. You can see the complex pseudo periodicity in the contour graphs in Kouznetsov's paper on the sexp_e.
- Sheldon

(12/04/2010, 11:09 PM)tommy1729 Wrote: a further remark and actually request is that i would love to see a plot of all the solutions , including branches
....

any volunteers ?

I am re-posting this reply after fixing a minor error. There is still some interest in the TommySexp solution (wikipedia talk), which as suggested previously, is infinitely differentiable, but probably nowhere analytic at the real axis. The goal of this post is to show the location of the singularities that limit the radius of convergence, for n=3 and n=4, to bolster the argument that the TommySexp function is nowhere analytic. The singularities occur wherever the 2sinh superfunction(z) has a value of 0 + 2nPi*I, where n is a positive integer.

Start with the n=3 case, renormalized with a bias=0.06783836607, so that TommySexp(0)=1. n=3 is sufficiently large to get more than double precision accurate results at the real axis, centered around z=0. Here is a plot of TommySexp(z) with a radius of 0.45, with z centered at the origin, for the n=3 case, which requires three logarithms.
For this plot with red=real, and green=imaginary. For this plot, n=3, with z=0.45*exp(I*theta), plotting TommySexp(z). Inside this region of the complex plane, the TommySexp function is analytic, with no singularities.
But now, we increase the sample radius merely 0.008, from 0.450 to 0.458. And now the TommySexp function becomes poorly behaved, because there are singularities inside the circle. Following this path to theta(Pi*I), the imag(TommySexp(-0.458 )) is no longer even zero, due to the singularities! This means the function's value has becomes dependent on which path around the singularity is chosen. Again, red is real, and green is imaginary.
What's going on, is that there are singularities, so that the TommySexp function has a limited radius of convergence. Next, I post the first fifty singularities of TommySexp, for the n=3 case. The fourth singularity is closest to the origin (after iterating three logarithms), with an absolute value of 0.457, which is why the second plot of the TommySexp(z), centered at the origin, with a radius=0.458 is poorly behaved.

The singularities occur wherever the 2sinh superfunction(z) has a value of 0 + 2nPi*I, where n is an integer>=1, because superf2sinh(z+1)=0, and the logarithm of zero is a singularity.
For an example, consider the fourth singularity on the list.
Let z= 2.285107737 + 0.3580209487*I
superf2sinh(z)=25.13274123*I=8*Pi*I.
superf2sinh(z+1)=0.

Now iterate the logarithm of superf2sinh(z+1), three times. We get a singularity on the first iteration. The logarithm, on a path circling the singularity gives different results, depending on the path. The logarithm of the logarithm of a singularity is also a singularity, which explains the second graph. All of these particular singularities line on a contour line, where real(superf2sinh(z))=0. So, now I plot this curve (red), superimposing on it a radius (z=0.458 ) half circle, showing how the sampling circle now includes the first singularity. This curve also shows that for the n=4 case, the radius of convergence is much much smaller still. It turns out the 600 thousandth singularity is at z=4.0025+0.0347i, so the radius of convergence for n=4 drops to 0.035. Using this definition, as n (number of iterated logarithms used to generate TommySexp) grows larger, the radius of convergence for TommySexp(z) at the origin gets arbitrarily small. Also, as the singularities approach the real axis, they also get arbitrarily close together. Thus, even though it can be shown that all of TommySexp's derivatives at the real axis are continuous, it is nonetheless probably nowhere analytic, just like the base change sexp definition discussed in an earlier post.
- Sheldon